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Centrifugal and centripetal force question: race-car and banked curve question |
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| Mar1-09, 07:53 AM | #1 |
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Centrifugal and centripetal force question: race-car and banked curve question
1. The problem statement, all variables and given/known data
A race-car driver is driving her car at a record-breaking speed of 225kh/h. The first turn on the course is banked at 15 degrees, and the car's mass is 1450kg. a) Calculate the radius of the curvature for this turn b) Calculate the centripetal acceleration of the car. c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction? d) What is the coefficient of static friction necessary to ensure the safety of this turn? 2. Relevant equations vector v^2 = (vector gravity)(radius)(tan incline of bank) centripetal acceleration = vector v ^2 / radius 3. The attempt at a solution First of all, I don't really understand centrifugal force. Centripetal force is for when an object has uniform circular motion, right? How is it affected by centrifugal force? So my basic understanding of this question is, when a car makes a turn it doesn't slide off laterally because of the friction between tires and the road, and if the curve has a bank it would increase force of friction when a car goes on it. Given: V = 225km/h = 225000m/h = 62.5m/s Incline of bank = 15 degrees M = 1450kg vector v^2 = (vector gravity)(radius)(tan incline of bank) So (radius) = (vector v)^2 / (vector gravity)(tank incline of bank) (radius) = (62.5m/s)^2 / (9.8m/s^2)(tan15) (radius) = (3906.25m^2/s^2) / (2.625902086m/s^2) (radius) = 1487.58m This is strange because if I divide (62.5m/s)^2 / (9.8m/s^2) first and then multiply it with (tan15) I get 106.80m So I'm not sure which answer is correct. I think all the other questions need a correct radius value so I can't go further :/ |
| Mar1-09, 08:11 AM | #2 |
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As far as this problem goes, you are not given enough information to solve it. Can you please tell me the textbook this is from and the problem number. (In case I have the textbook.) |
| Mar1-09, 08:28 AM | #3 |
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What other information do I need to solve it? |
| Mar1-09, 08:39 AM | #4 |
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Centrifugal and centripetal force question: race-car and banked curve question |
| Mar2-09, 09:06 PM | #5 |
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Can i sue them for giving flawed questions?
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| Mar15-09, 11:48 AM | #6 |
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you better read the contents before you go to this question.
firstly, you have to draw a FBD. a)Fn=Fg/cos15 Fn=14711N Fc=sin15 Fc=14711*sin15 Fc=3807N r=Fc/mv^2 r=6.72*10^4m b) a=Fc/m a=2.626m/s^2 |
| Mar15-09, 01:17 PM | #7 |
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(And your equation for r is upside down.) |
| May13-10, 07:34 PM | #8 |
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ok so i just did that 2 weeks ago, got full marks for it XD.
so for a) u use v= sqrt ( g r tan(angle)) v= velocity g= earth's gravitational force v= 225km/h = 62.5 m/s angle= 15 m= 1450kg so i'm not going to draw a FBD, you do that yourself, now continueing with a) v^2= g r tan(angle) 62.5^2=9.8 r tan 15 r=3906.25 / 2.63 r=1485.3m b) ac = centripetal acceleration ac= v^2 / r , meh i'm not going to write the rest of the problem to this because i believe you can finish this. solution = 2.63 m/s^2 c) Fs max = m g sin(angle) Fs max= magnitude of the force of static friction solution = 3677.8N d) (can't make the mu symbol for coefficient of friction >< ) mu s = tan (angle) mu s = tan 15 mu s = 0.268 you're very welcome, please send me a fruit basket for x-mas =D |
| May13-10, 07:56 PM | #9 |
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Sorry, no fruit basket. |
| May13-10, 08:05 PM | #10 |
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darn it i really wanted that fruit basket XD .
well i say we should complain to the teachers, after all im only the poor little student that is using the equations that are given in the book >.> . but i did get 3/3 marks for that question, therefore i assume it was meant to be a flawed example for the simple purpose of teaching the student? |
| May13-10, 08:12 PM | #11 |
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| May13-10, 09:38 PM | #12 |
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yes i understand how the equation was obtained, but i'm in the process of learning so i might think i know but i might not know at all.
anyways what books on classical physics for around grade 12 and perhaps beyond would you recommend that explain it well and are reasonable to understand?? |
| May14-10, 01:20 PM | #13 |
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Here are some web resources: A Banked Turn - No Friction & A Banked Turn With Friction |
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